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$xhtml = array(
	'<{title}>' => 'Polynomials',
	'<{subtitle}>' => 'Written in <span title="College Algebra">MATH 1201</span> by <a href="https://y.st./">Alex Yst</a>, finalised on 2018-02-21',
	'<{copyright year}>' => '2018',
	'takedown' => '2017-11-01',
	'<{body}>' => <<<END
<section id="problem0">
	<h2>Problem 0</h2>
	<p>
		A seventh-degree polynomial is considered to be &quot;odd&quot;, as the power of the leading term is odd.
		&quot;Odd&quot; polynomials always eventually continue in one direction in a positive manner and in the other in a negative manner.
		As a result, <strong>an &quot;odd&quot; polynomial (such as a seventh-degree polynomial) will always have at least one zero</strong>.
		The fundamental theorem of algebra also states that if a polynomial of degree greater than one has complex coefficients, there is always at least one complex zero.
		As a complex zero is a zero, there must be at least one zero.
		In theory, there are always exactly seven zeroes for a seventh-degree polynomial, but those zeros might not be necessarily distinct; that is, the polynomial may have multiple of the same zero.
	</p>
	<p>
		Intuitively, a polynomial may cross the x-axis up to once for each degree.
		The textbook confirms this.
		According to Theorem 3.7, an n-degree polynomial may have as many as n zeros, so <strong>a seventh-degree polynomial may have up to seven zeros</strong>.
	</p>
	<p>
		It&apos;s worth noting that the question states that at least one root of the seventh-degree polynomial is a complex number.
		The book defines a complex number as being a number in the form a + bi, where i is the imaginary unit and a, b, both, or neither could be zero.
		What this tells us is that the number zero, which could be written as &quot;0 + 0i&quot; is a complex number.
		In fact, every real number is a complex number; not all complex numbers are imaginary.
		This means that the fact that the hypothetical seventh-degree polynomial has a complex root is irrelevant, and in fact, <strong>*all*</strong> roots of <strong>*all*</strong> polynomials are complex.
	</p>
</section>
<section id="problem1">
	<h2>Problem 1</h2>
	<p>
		Using the synthetic division proposed by the textbook, we can easily divide x<sup>8</sup> - 1 by x - 1:
	</p>
<pre>
1 | 1 0 0 0 0 0 0 -1
    ⭣ 1 1 1 1 1 1  1
   _________________
    1 1 1 1 1 1 1  0</pre>
	<p>
		Using that method, we see that (x<sup>8</sup> - 1) / (x - 1) is equal to <strong>x<sup>7</sup> + x<sup>6</sup> + x<sup>5</sup> + x<sup>4</sup> + x<sup>3</sup> + x<sup>2</sup> + x</strong>.
	</p>
</section>
<section id="problem2">
	<h2>Problem 2</h2>
	<p>
		To find the zeros of x<sup>3</sup> + 5x<sup>2</sup> - 7x + 1 seems like an impossible task.
		However, that&apos;s only because the assigned reading material for the week was sections 3.1, 3.2, and 3.4, while a vital piece of the information we need for this sort of problem is in the skipped section, section 3.3.
		Section 3.3 gives us a great lead as to where to look for the first zero to factor out.
		Using the rational zeros theorem found there, we see that the possible rational zeros for the equation are x = ±1.
		-1 doesn&apos;t work in the division, so it&apos;s not a zero, but 1 works as follows:
	</p>
<pre>1 | 1  5 -7  1
    ⭣  1  6 -1
  ____________
    1  6 -1  0</pre>
	<p>
		That gives us the factorization (x - 1)(x<sup>2</sup> + 6x - 1).
		The rational zeros theorem tells us no other rational zeros are possible, so further factoring would be very difficult by normal means, if not outright impossible.
		However, we&apos;ve managed to get the degree down to two, making it a quadratic equation.
		We can thus use the quadratic formula to find the other zeros:
	</p>
	<p>
		0 = (-6 ± √(6<sup>2</sup> - 4 * 1 * -1)) / (2 * 1)<br/>
		0 = (-6 ± √(36 + 4)) / 2<br/>
		0 = (-6 ± √40) / 2<br/>
		0 = (-6 ± 2√10) / 2<br/>
		0 = (-6 / 2) ± (2√10 / 2)<br/>
		0 = -3 ± √10
	</p>
	<p>
		This means the three zeros are <strong>1 and -3 ± √10 (1, -3 + √10, and -3 - √10)</strong>
	</p>
</section>
END
);

